Let $R$ be a commutative ring with identity, and let $S \subseteq R$ be a subring of $R$, sharing a common identity. Moreover, let $M$ and $N$ be $R$-modules. Is it then true that $M \otimes_R N \cong M \otimes_S N$ as $S$-modules?
It feels like I can just define a function $f : M \otimes_R N \to M \otimes_S N$ by $f(a \otimes b) = a \otimes b$, which would be bijective and $S$-linear. However, my intuition says that the isomorphism should not hold generally.
Actually the map $a\otimes b\mapsto a\otimes b$ is well-defined $M\otimes_SN\to M\otimes_RN$ but not the other way. In particular, we have $ra\otimes b=a\otimes rb$ in $M\otimes_RN$ but not $M\otimes_SN$ in general (for $r\in R\setminus S$).
In general, $M\otimes_SN$ is "bigger" because you're quotienting by fewer relations than $M\otimes_RN$.
For example, consider $S=\mathbb{F}_2$ and $R=\mathbb{F}_2[\varepsilon]/(\varepsilon^2)$. Then as $\mathbb{F}_2$-vector spaces, $R\otimes_SR$ has dimension $4$ whereas $R\otimes_RR$ has dimension $2$, so they can't be isomorphic.